Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{7(5r + 2)}{r} \times \dfrac{8r}{4(5r + 2)} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ 7(5r + 2) \times 8r } { r \times 4(5r + 2) } $ $ z = \dfrac{56r(5r + 2)}{4r(5r + 2)} $ We can cancel the $5r + 2$ so long as $5r + 2 \neq 0$ Therefore $r \neq -\dfrac{2}{5}$ $z = \dfrac{56r \cancel{(5r + 2})}{4r \cancel{(5r + 2)}} = \dfrac{56r}{4r} = 14 $